Question

The effective capacitance between P and Q of the figure below is :

$C_1=C_3=C_4=C_5=10\mu F$

$C_2=100\mu F$

• A. $10\mu F$
• B. $40\mu F$
• C. $5\mu F$
• D. $20\mu F$

Answer 1

The correct option is A $10\mu F$

Equivalent wheat stone bridge capacitors circuit for given circuit is shown in the fig.

In given circuit,

$\frac{C_1}{C_3}=\frac{10}{10}=1$ and $\frac{C_4}{C_5}=\frac{10}{10}=1$

$\therefore \frac{C_1}{C_3}=\frac{C_4}{C_5}$

$\therefore$ By wheat stone bridge principle there will be no p.d. across $C_2$ of $100\mu F$ hence we can remove $C_2$.

$\therefore$ Effective capacitance between $P$ and $Q$ is given by,

$C_{PQ}=\frac{C_1{C}_4}{C_1+C_4}+\frac{C_3{C}_5}{C_3+C_5}$

$C_{PQ}=\frac{100}{10+10}+\frac{100}{10+10}=5+5=10\mu F$